Everyone, about the space of sequences $S$; suppose the distance function map the sequence to a real number
e.g. $d(a,b)=\sup_n|a_n-b_n|$ for $n \in \mathbb{N}$
I am confusing about this concept: do we consider each sequence as an element, like a function of $n \in \mathbb{N}$, or do we consider each item in sequences as an element?
I figure it should be the first case, then I am not sure how to deal with the proof of completeness?
let $S$ be a space of sequences such that $a=\{a_n\}_{n=1}^\infty$. Suppose all sequences obey $\lim_n |a_n|=0$.
then can I create a sequence of function with $f_1=a_1$ (the first sequence in the space ... $f_n=a_n$ ... ). I can show that this sequence of functions converges to an $f$, and then I am not sure; can I use the Cauchy criterion to say that some $a_k \in S$ satisfies to be the $f$, as it is an elment of $S$, then $S$ is complete?
I figure that above is probably not right, anyway, my question is how to show $S$ is complete.
Hope I made my question clear.
Thank you!
You did not clearly define the set $S$. I am assuming it is $$ S=\left\{ f:\mathbb N\to \mathbb R \vert \lim_{n\to \infty} f(n)=0\right \} $$ with a metric $$ d(f,g)=\sup_{n\in\mathbb R}\vert f(n)-g(n) \vert. $$
Ansatz to proof that $(S,d)$ is complete: Consider a cauchy sequence $f^1,f^2,\dots \in S$. Then $$d(f^{n},f^{m})>= \vert f^{n}(k)-f^{m}(k) \vert $$ so that $n\mapsto f^n(k)$ is a cauchy sequence in $\mathbb R$ for every $k\in \mathbb N$. Since $\mathbb R$ is complete there exists some $f(k)\in\mathbb R$ such that $\lim_{n\to \infty}f^n(k)=f(k)$. Now prove that $\lim_{n\to \infty} f(n)=0$ and $\lim_{n\to \infty} f^n=f$.